CF510B Fox And Two Dots(搜索图形环)
2 seconds
256 megabytes
standard input
standard output
Fox Ciel is playing a mobile puzzle game called "Two Dots". The basic levels are played on a board of size n × m cells, like this:
Each cell contains a dot that has some color. We will use different uppercase Latin characters to express different colors.
The key of this game is to find a cycle that contain dots of same color. Consider 4 blue dots on the picture forming a circle as an example. Formally, we call a sequence of dots d1, d2, ..., dk a cycle if and only if it meets the following condition:
- These k dots are different: if i ≠ j then di is different from dj.
- k is at least 4.
- All dots belong to the same color.
- For all 1 ≤ i ≤ k - 1: di and di + 1 are adjacent. Also, dk and d1 should also be adjacent. Cells x and y are called adjacent if they share an edge.
Determine if there exists a cycle on the field.
The first line contains two integers n and m (2 ≤ n, m ≤ 50): the number of rows and columns of the board.
Then n lines follow, each line contains a string consisting of m characters, expressing colors of dots in each line. Each character is an uppercase Latin letter.
Output "Yes" if there exists a cycle, and "No" otherwise.
3 4
AAAA
ABCA
AAAA
Yes
3 4
AAAA
ABCA
AADA
No
4 4
YYYR
BYBY
BBBY
BBBY
Yes
7 6
AAAAAB
ABBBAB
ABAAAB
ABABBB
ABAAAB
ABBBAB
AAAAAB
Yes
2 13
ABCDEFGHIJKLM
NOPQRSTUVWXYZ
No
In first sample test all 'A' form a cycle.
In second sample there is no such cycle.
The third sample is displayed on the picture above ('Y' = Yellow, 'B' = Blue, 'R' = Red).
【题意】
给定n*m矩阵,看是否有相同的字符可以构成一个环
【分析】
爆搜~
注意:1、构成环至少需要4个字符
2、注意判断字符的来路
【代码】
#include<cstdio>
#include<cstdlib>
using namespace std;
const int N=105;
int n,m,ans,dir[4][2]={{0,1},{0,-1},{1,0},{-1,0}};
char mp[N][N];bool vis[N][N]={0};
void dfs(int x,int y,int px,int py,int step){
vis[x][y]=1;
for(int i=0;i<4;i++){
int nx=x+dir[i][0];
int ny=y+dir[i][1];
if(nx<1||ny<1||nx>n||ny>m||mp[nx][ny]!=mp[x][y]) continue;
if(!vis[nx][ny]) dfs(nx,ny,x,y,step+1);
else{
if((nx!=px||ny!=py)&&step>=4){puts("Yes");exit(0);}
}
}
}
inline void Init(){
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++) scanf("%s",mp[i]+1);
}
inline void Solve(){
for(int i=1;i<=n;i++){
for(int j=1;j<=m;j++){
if(!vis[i][j]){
dfs(i,j,0,0,1);
}
}
}
puts("No");
}
int main(){
Init();
Solve();
return 0;
}
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